Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int setBit(int N, int pos) { return N = N | (1 << pos); }
inline int resetBit(int N, int pos) { return N = N & ~(1 << pos); }
inline bool checkBit(int N, int pos) { return (bool)(N & (1 << pos)); }
const int MAX = 105;
int n, m, arr[MAX][MAX], save[MAX][MAX];
inline bool checkRow(int idx) {
  int i;
  for (i = 0; i < m; i++) {
    if (arr[idx][i] < 1) return false;
  }
  return true;
}
inline bool checkCol(int idx) {
  int i;
  for (i = 0; i < n; i++) {
    if (arr[i][idx] < 1) return false;
  }
  return true;
}
inline bool check() {
  int i, j;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (arr[i][j] != 0) return false;
    }
  }
  return true;
}
inline vector<pair<int, int> > solveRow() {
  int i, j;
  vector<pair<int, int> > v;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      arr[i][j] = save[i][j];
    }
  }
  while (true) {
    bool flag = true;
    for (i = 0; i < n; i++) {
      if (checkRow(i)) {
        for (j = 0; j < m; j++) arr[i][j]--;
        v.push_back({0, i});
        flag = false;
        break;
      }
    }
    if (flag) {
      for (i = 0; i < m; i++) {
        if (checkCol(i)) {
          for (j = 0; j < n; j++) arr[j][i]--;
          v.push_back({1, i});
          flag = false;
          break;
        }
      }
    }
    if (flag) break;
  }
  if (check())
    return v;
  else {
    v.clear();
    v.push_back({-1, -1});
    return v;
  }
}
inline vector<pair<int, int> > solveCol() {
  int i, j;
  vector<pair<int, int> > v;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      arr[i][j] = save[i][j];
    }
  }
  while (true) {
    bool flag = true;
    for (i = 0; i < m; i++) {
      if (checkCol(i)) {
        for (j = 0; j < n; j++) arr[j][i]--;
        v.push_back({1, i});
        flag = false;
        break;
      }
    }
    if (flag) {
      for (i = 0; i < n; i++) {
        if (checkRow(i)) {
          for (j = 0; j < m; j++) arr[i][j]--;
          v.push_back({0, i});
          flag = false;
          break;
        }
      }
    }
    if (flag) break;
  }
  if (check())
    return v;
  else {
    v.clear();
    v.push_back({-1, -1});
    return v;
  }
}
inline void print(vector<pair<int, int> > v) {
  printf("%d\n", (int)v.size());
  for (auto x : v) {
    if (x.first == 0)
      printf("row ");
    else
      printf("col ");
    printf("%d\n", x.second + 1);
  }
}
int main() {
  int i, j;
  scanf("%d %d", &n, &m);
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      scanf("%d", &arr[i][j]);
      save[i][j] = arr[i][j];
    }
  }
  auto a = solveRow();
  auto b = solveCol();
  bool fa = true, fb = true;
  if ((int)a.size() == 1 && a[0].first == -1) fa = false;
  if ((int)b.size() == 1 && b[0].first == -1) fb = false;
  if (!fa && !fb)
    puts("-1");
  else if (fa && !fb)
    print(a);
  else if (!fa && fb)
    print(b);
  else {
    if ((int)a.size() <= (int)b.size())
      print(a);
    else
      print(b);
  }
  return 0;
}